Electrostatic Potential And Capacitance

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Question
CBSEENPH12040111

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

  • Increases by a factor of 4

  • Decreases by a factor of 2

  • Increases by a factor of 2

  • Remains the same

Solution

B.

Decreases by a factor of 2

WiredFaculty

Charge on capacitor
q = CV
when it is connected with another uncharged capacitor.
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Question
CBSEENPH12039778

A capacitor of 2μF is charged as shown in the figure. When the switch s is turned to position 2, the percentage of its stored energy dissipated is,

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  • 20%

  • 75%

  • 80%

  • 0%

Solution

C.

80%

Consider the figure given above.
When switch S is connected to point 1, then initial energy stored in the capacitor is given as,
straight E space equals space 1 half open parentheses 2 μF close parentheses xV squared
When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,
WiredFaculty
Therefore, per centage loss of energy = fraction numerator 1.6 over denominator 2 end fraction x 100 space equals space 80 percent sign

Question
CBSEENPH12040104

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting thecharging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates

  • decreases

  • does not change

  • becomes zero

  • increases

Solution

D.

increases

Charge remains constant after charging. 
If the battery is removed after charging then the charge stored in the capacitor remains constant. 
                q = constant
Change in capacitance
                  straight C apostrophe space equals space fraction numerator straight E subscript 0 straight A over denominator straight d apostrophe end fraction
As                straight d apostrophe greater than straight d
Hence,        straight C apostrophe less than straight C
Hence, potential difference between the plates
                       straight V apostrophe space equals space fraction numerator straight q over denominator straight C apostrophe end fraction
or      DV apostrophe space proportional to space space fraction numerator 1 over denominator straight C apostrophe end fraction
As capacitance decreases, so potential difference increases.

Question
CBSEENPH12039797

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of a dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. which of the following is incorrect?

  • The potential difference between the plates decreases K times

  • The energy stored in the capacitor decreases K times

  • The change in energy stored is 1 half C V squared space open parentheses 1 over K minus 1 close parentheses

  • The charge on the capacitor is not conserved

Solution

C.

The change in energy stored is 1 half C V squared space open parentheses 1 over K minus 1 close parentheses

WiredFaculty
As the battery is disconnected from the capacitor the charge will not be destroyed i.e. q' = q with the introduction of dielectric in the gap of capacitor the new capacitance will be
C' = CK
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Question
CBSEENPH12039821

A parallel plate capacitor has a uniform electrinc field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy storedf in the capacitor is df and area of each plate  is A, the energy stored in the capacitor is 

  • 1 half straight epsilon subscript straight o straight E squared
  • straight E squared Ad divided by straight epsilon subscript straight o
  • 1 divided by 2 space straight epsilon subscript straight o straight E squared Ad
  • straight epsilon subscript straight o EAd

Solution

C.

1 divided by 2 space straight epsilon subscript straight o straight E squared Ad

 Energy density for a parallel plate capacitor
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